Find the equation of the circle passing through the points of intersection of the circle $x^{2} + y^{2} - 6 x + 2 y + 4 = 0$ and $x^{2} + y^{2} + 2 x - 4 y - 6 = 0$ and with its centre on the line $y = x$

x^{2} + y^{2} + \frac{10 x}{7} - \frac{10 y}{7} + \frac{12}{7} = 0

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x^{2} + y^{2} - \frac{10 x}{7} + \frac{10 y}{7} + \frac{12}{7} = 0

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x^{2} + y^{2} + \frac{10 x}{7} + \frac{10 y}{7} - \frac{12}{7} = 0

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x^{2} + y^{2} - \frac{10 x}{7} - \frac{10 y}{7} - \frac{12}{7} = 0

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Solution

The correct option is B $x^{2} + y^{2} - \frac{10 x}{7} - \frac{10 y}{7} - \frac{12}{7} = 0$
equation of the circle passing through the points of intersection of the circle $x^{2} + y^{2} - 6 x + 2 y + 4 = 0$ and $x^{2} + y^{2} + 2 x - 4 y - 6 = 0$ is $x^{2} + y^{2} - 6 x + 2 y + 4 + k (x^{2} + y^{2} + 2 x - 4 y - 6) = 0$
Center is $(\frac{3 - k}{1 + k}, \frac{- 1 + 2 k}{1 + k})$
Since, it lie on $y = x$
Therefore, $\frac{3 - k}{1 + k} = \frac{- 1 + 2 k}{1 + k}$
$\Rightarrow k = \frac{4}{3}$
Therefore, $x^{2} + y^{2} - \frac{10 x}{7} - \frac{10 y}{7} - \frac{12}{7} = 0$
Ans: D

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Find the equation of the circle passing through the points of intersection of the circle x2+y2−6x+2y+4=0 and x2+y2+2x−4y−6=0 and with its centre on the line y=x

Find the equation of the circle passing through the points of intersection of the circle $x^{2} + y^{2} - 6 x + 2 y + 4 = 0$ and $x^{2} + y^{2} + 2 x - 4 y - 6 = 0$ and with its centre on the line $y = x$